Question

The area bounded by the curves $y={(x-1)}^2,y={(x+1)}^{2}andy=\frac{1}{4}$ is

• A. $\frac{1}{3}$
• B. $\frac{2}{3}$
• C. $\frac{1}{4}$
• D. $\frac{1}{5}$

Answer 1

The correct option is A

$\frac{1}{3}$

Explanation for correct option

Finding the area

By solving the equations we get,

Points of intersection are:

$\begin{array}{rcl}{(x-1)}^2& =& \frac{1}{4}\\ & \Rightarrow & x=\frac{1}{2}\end{array}$

Also,

$\begin{array}{rcl}{(x+1)}^2& =& \frac{1}{4}\Rightarrow x=\frac{-1}{2}\end{array}$

i.e.,$Q\left(\frac{1}{2},\frac{1}{4}\right)andR\left(-\frac{1}{2},\frac{1}{4}\right)$

Therefore, required area is:

$\begin{array}{rcl}A& =& {\int }_{-\frac{1}{2}}^0\left({(x+1)}^2-\frac{1}{4}\right)dx+{\int }_0^{\frac{1}{2}}\left({(x-1)}^2-\frac{1}{4}\right)\\ & =& {\left[\frac{{(x+1)}^3}{3}-\frac{x}{4}\right]}_{-\frac{1}{2}}^0+{\left[\frac{{(x-1)}^3}{3}-\frac{x}{4}\right]}_0^{\frac{1}{2}}\left({\int }_a^b{\left(x+p\right)}^n={\left[\frac{{\left(x+p\right)}^{n+1}}{n+1}\right]}_a^b\right)\\ & =& \left[(\frac{1}{3}-0)-\left(\frac{1}{24}+\frac{1}{8}\right)\right]+\left[\left(-\frac{1}{24}-\frac{1}{8}\right)-\left(\frac{-1}{3}-0\right)\right]\\ & =& \frac{1}{6}+\frac{1}{6}\\ & =& \frac{1}{3}\end{array}$

Hence, option (A) is the correct answer.